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2 years ago

a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r

Physics

1 answer:

dimaraw [331]2 years ago

4 0

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

spring constant k= 2250 Nm

now according to principal of conservation of energy we observe,

the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.

mgh= μ (mgl) +1/2 kx²

10 x 10 x 3= μ(600) +(1125) (0.09)

μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

brainly.com/question/13754413

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A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th

-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = -13.125 m/s^2$

Explanation:

By momentum conservation along x direction we will have

$mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2$

so we have

$v_1 + v_2 = 2v$

$v_1 + v_2 = 4.68$

also by energy conservation

$\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J$

$\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17$

$(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)$

$(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83$

$21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83$

$2v_2^2 - 9.36v_2 + 2.12 = 0$

by solving above equation we will have

$v_1 = 4.44 m/s$

$v_2 = 0.24 m/s$

Average acceleration on first part of the chunk is given as

$a_1 = \frac{4.44 - 2.34}{0.16}$

$a_1 = 13.125 m/s^2$

Average acceleration on second part of the chunk is given as

$a_2 = \frac{0.24 - 2.34}{0.16}$

$a_2 = -13.125 m/s^2$

5 0

4 years ago

Lightning flashes one mile (1609 m) away from you. How much time does it take the light to travel that distance?

hichkok12 [17]

Answer:

$t=5.36\times 10^{-6}\ s$

Explanation:

Given that,

Lightning flashes one mile (1609 m) away from you.

We need to find the time it take the light to travel that distance. Let the time be t. We know that,

speed = distance/time

$t=\dfrac{d}{v}\\\\t=\dfrac{1609}{3\cdot10^{8}}\\\\=5.36\times 10^{-6}\ s$

So, the required time is $5.36\times 10^{-6}\ s$

7 0

3 years ago

For crystal diffraction experiments, wavelengths on the order of 0.25 nm are often appropriate.

Kamila [148]

Answer:

A) E = 4.96 x 10³ eV

B) E = 4.19 x 10⁴ eV

C) E = 3.73 x 10⁹ eV

Explanation:

For photon energy is given as:

$E = hv$

$E = \frac{hc}{\lambda}$

where,

E = energy of photon = ?

h = 6.625 x 10⁻³⁴ J.s

λ = wavelength = 0.25 nm = 0.25 x 10⁻⁹ m

Therefore,

$E = \frac{(6.625 x 10^{-34} J.s)(3 x 10^8 m/s)}{0.25 x 10^{-9} m}$

$E = (7.95 x 10^{-16} J)(\frac{1 eV}{1.6 x 10^{-19} J})$

E = 4.96 x 10³ eV

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of electron = ?

m₀ = rest mass of electron = 9.1 x 10⁻³¹ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (9.1 x 10^{-31} kg)(3 x 10^8 m/s)^2\\$

$E = (8.19 x 10^{-14} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

E = 4.19 x 10⁴ eV

The energy of a particle at rest is given as:

$E = m_{0}c^2$

where,

E = Energy of alpha particle = ?

m₀ = rest mass of alpha particle = 6.64 x 10⁻²⁷ kg

c = speed of light = 3 x 10⁸ m/s

Therefore,

$E = (6.64 x 10^{-27} kg)(3 x 10^8 m/s)^2\\$

$E = (5.97 x 10^{-10} J)(\frac{1 eV}{1.6 x 10^{-19} J})\\$

E = 3.73 x 10⁹ eV

8 0

3 years ago

Solve:40KMEN

Oksanka [162]

The spider will cross the driveway 25.71 in seconds.

Why?

It's a conversion problem, so, in order to solve it, we need to convert the units of the given information.

Let's convert the given speed (in cm/s) to m/s.

We know that:

$1m=100cm$

So, converting we have:

$14\frac{cm}{s}*\frac{1m}{100cm}=0.14\frac{m}{s}$

Then, calculating the time, we have:

$distance=speed*time\\\\3.6m=0.14\frac{m}{s}*time\\\\time=\frac{3.6m}{0.14\frac{m}{s}}=25.71s$

We have that the spider will cross the driveway in 25.71 seconds.

Have a nice day!

8 0

4 years ago

A baseball leaves a bat with a horizontal velocity of 20.0 m/s. When it has left the bat for a time of 0.250 s (Assume air resis

miss Akunina [59]

Answer:

The distance is 5 m.

Explanation:

Given that,

Horizontal velocity = 20.0 m/s

Time t = 0.250 s

We need to calculate the horizontal distance

The distance traveled by the baseball with horizontal velocity in time t is given by

Using formula for horizontal distance

$d = v\times t$

$d = 20.0\times0.250$

$d=5\ m$

Hence, The distance is 5 m.

5 0

3 years ago